**
QUESTION:***
*
My dad told me about your website, very interesting reading. My question deals with molecules. When a molecule emits a photon, the mass of the molecule decreases to account for the energy in the photon. So, the mass of the molecule as a whole decreases, but this mass does not come from the "parts" of the molecule. In other words, the mass of the constituent electrons does not decrease, the mass of the protons does not decrease, so the energy must come from the electric field between the electrons and protons.
But the electric field has energy, not mass. Now mass is a form of energy, but I don't think you can say that the field has mass? But yet, it is said that the mass of the molecule decreases. The electric field contributes to the mass of the molecule, but yet it is incorrect to say that the field has mass?

**ANSWER: **

Actually, this is not as complicated as you are trying to make it. It
all boils down to the fact that mass *is* a form of energy and must be factored into
any energy conservation that occurs in an isolated system. You say that
the masses of the protons and electrons do not change, but that is not
right. Look at the simplest case, a hydrogen atom. If you measure the
mass of this atom it will be less than if you measure the masses of a
free electron and a free proton. Here is how you can see that: if you
pull the electron away from the proton, that is you ionize the atom, do
you have to do any work? Of course you do because the electron and
proton are bound together. So, you have added energy to the system (p+e)
and that energy shows up as mass. In a system as complicated as a
molecule you cannot say which particle or particles changed their
masses, but you can say for sure that the total mass of the molecule
changed by exactly the energy of the emitted photon divided by *c*^{2}.

**
QUESTION:***
*
This is my 2nd question related to the issue of time dilation - this one being related to the issue of motion [the other being based on gravity]. Since time dilation occurs for all moving objects, and considering further that the Earth has been revolving around the sun at 30 km per second for the last 4 billion years -- and further that our solar system is moving at roughly 45K mph through space, can't it be said that, compared to other objects in the universe, time dilation has occurred to a significant degree for our planet over those 4 billions years? And that really every object in the universe likewise has its own unique time dilation associated with it? Can't it also be said that every consolidated arrangement of matter in the universe is moving along at different "rates of time?" Wouldn't, over the course of several billion years, these "pockets" of different time spans become more and more "incompatible" with each other?

**ANSWER: **

The two speeds you quote are about the same (45,000 mph≈2x10^{4}
m/s and 30x10^{3} km/s=3x10^{4} m/s). So let's just
choose the larger one and see how much time dilation there is. Relative
to the sun, an elapsed time *T*=4x10^{9} y would
correspond to *T'=ΥT *wwhere *Υ=*1/√(1-(*v*/*c*)^{2})=1/√(1-(3x10^{4}/3x10^{8})^{2})≈(1+0.5x10^{-8}).
Therefore *T'≈*(*T*+20), a difference of 20 years. This
may sound like a pretty long time to you, but relative to 4 billion it
is less than 10^{-6} %. That is not "significant" to my mind.
You are right that every object has its own clock which, relative to
other clocks, is not necessarily the same; every object also has its own
meter stick, not necessarily the same as other meter sticks in the
universe. The important thing is that you always must talk about
velocity with respect to what.

**
QUESTION:***
*
I have a question about light please. I was thinking about one of Einstein's thought experiments. Let me set some the situation first - there is a stationary observer at a train station. Einstein is on a train travelling at 99.99999%c and is about to pass the train station. As the front of the train reaches the train station, the stationary person emits a light beam in the same direction as the train while at the same time, the train turns on its headlight. Based on what I understand, both beams of light should be travelling at C but I am interested in understanding what each person sees. I can understand that the stationary person will see both light beams leave him at C, along with the train at 99.99999%c. But what about Einstein on the train? I understand that you cannot add the velocity of the train to its headlight beam. But I can imagine that he would see the beams moving ahead and away from him at .00001%c since he is already travelling at 99.99999%c. But that seems to violate the idea C being the same for everyone in every reference frame. I can also imagine that due to that rule, he would see both light beams moving ahead and away from him at C. But if that's true, how can he possibly be going 99.99999%c if the beams are moving away from him at C and how could the stationary person see both the light moving at C and Einstein moving at 99.99999%c? I need your insight to understand this please.

**ANSWER: **

This is a very long question, but the answer is very short. All
observers, regardless of their motions, see the speed of any
electromagnetic radiation to be *c*. Your statement that the
train rider would see the beams having is speed 0.00001*c* is
simply wrong. That is not to say that the beams are unchanged—they
would be Doppler-shifted to different wavelengths, but still move with
speed *c*.
See the faq page to help yourself to accept that this is true.

**
FOLLOWUP QUESTION:***
*
Thanks! I do accept what you are saying is true. Acceptance was never the issue, understanding was.
Since the doppler is shifted then what Einstein would see with his eyes would be the light around him becoming redder and redder the closer to C he gets?

**ANSWER: **

I hope you did go to the faq page and now have an "understanding"; if
not, you should look at these links
there. You might also read an
earlier answer about relativistic velocity addition. You are right,
if light with frequency *f *is moving in the same direction as
you are, you will see a lower frequency light *f'*,
red-shifted. The Doppler equation is *f'=f*√[(1-*β*)/(1+*β*)] where *β=v*/*c. *
Suppose we take your case where *β=*0.9999999=10^{-7}
so √[(1-*β*)/(1+*β*)]=2.24x10^{-4} and choose *λ*=550
nm (yellow light). Then *f=c*/*λ*=3x10^{8}/5.5x10^{-7}=5.45x10^{14}
Hz and* f'*=2.24x10^{-4}x5.45x10^{14}=1.23x10^{11}
Hz=123 GHz. This frequency is in the short-wave radio range, nothing he could "…see
with his eyes…"! The Doppler effect has red-shifted the wave
way beyond visible red!

**
QUESTION:***
*
Dear Physicist, my question relates to special relativity and time dilation. This question may sound dumb but it continues to bother me. I have a space ship which I am sending to Alpha Centauri, the distance is 4.3 light years. I am able to push the ship to .7 the speed of light.
According to my understanding of time dilation, events on the ship will slow down due to the increase in the mass put into the system because, I am approaching the speed of light (a fixed constant) is it correct events back on Earth, which is moving at a much lower percent of the speed of light, will pass at a faster rate? If so, then doesn't that mean increasing the velocity of the ship to get there faster is really self defeating because instead of reaching Alpha Centauri in 5.59 years, time dilation will cause the actual time measured in Earth time to be much longer?

**ANSWER: **

I am afraid you have some serious misconceptions. Time dilation says
that if you are moving with respect to me, I observe your clock to
run slowly. Do not mistake that
for saying that your clock *appears*
to run slowly—it might
appear to run faster or slower
than mine, but is running slowly. Furthermore, it has nothing to do with
mass increasing with speed. But how do you observe things? Your clock
appears to run just the same as if you were at rest; but wait, you are
at rest relative to yourself and you see me moving and therefore find
that my clock runs slowly! Now, the time that I see you getting to the
star is 4.3/0.7=6.14 years. But, you see it as a considerably shorter
time. How can that be if your clock is running normally and you clearly
see your speed as 0.7*c*? The answer is length contraction—you
observe the distance to the star to be less than 4.3 light years by the
gamma factor, √(1-0.7^{2})=0.71, or 3.1 light years so the
time will be 3.1/0.7=4.43 years. If you could get going a speed 0.9*c*,
you could get there in less than half a year by your clock but still
6.14 by mine. I have given you a couple of links to the
faq page, but you might want to peruse that page
for more relativity answers which might interest you, particularly the
twin paradox.

**QUESTION:***
*
If any particle drop falling from height h, then velocity increases but in case of photon with light veloctiy c falling drop then what changes occurs, it will increase velocity or not?
If velocity of photon increase then how its possible because no one have
velocity is greater then speed of light?

**ANSWER: **

A photon gains energy when it "falls" in a gravitational field, just
like a rock does. But, it does not speed up because the speed of light
is a universal constant. The energy of a photon is proportional to its
frequency, so as its energy increases its frequency increases. As light
travels (with constant speed) in the direction of a gravitational field
(down) it shifts toward the blue end of the spectrum and this is called
a blue shift. As light travels (with constant speed) opposite the
direction of a gravitational field (up) it shifts toward the red end of
the spectrum and this is called a red shift.

**QUESTION:***
*
Two frames xOy and x'O'y' are in uniform motion along their x axes. We will
consider for simplicity the first frame to be "fixed" while the second one
moves to the right with a velocity v. From the origin of the moving frame
x'O'y' two rays of light are emitted simultaneously, one along the axis O'x'
and the other one at an angle of 60 degrees with the O'x' axis. Two mirrors
are placed at the same distance l on the two tracks and the light gets
reflected. Obviously, the two reflected rays, as observed in the frame x'Oy'
return to 0' at the same time. I made some calculations for v = c/2 and l =
1,000,000 m. Surprisingly, the two rays, as observed from frame xOy, do not
appear to meet return to the origin at the same time. I can share my
calculations with you. It is possible that I made a mistake. However if my
calculations are correct, this would be a very strange thing indeed.

**ANSWER: **

First of all, it would not be "…a
very strange thing indeed…" to find what is
simultaneous in one frame is not in another. One of the keystones of
special relativity is that simultaneaty of two events depends on the
frame of reference. I have not included your solution to the question
because I prefer to work it out myself rather than troubleshoot your
work. The figure shows the situation as seen by each observer. The
primed system moves in the +*x* direction while the red-drawn
distances to the mirrors are at rest in the unprimed system. The rest
lengths of the two distances are *L* and the angle one makes with
the *x*-axis is *θ*. The round-trip time along the
*x*-axis is *t*_{1}=2*L*/*c* and
along the other length is the same, *t*_{2}=2*L*/*c*.
In the moving system all lengths along the the *x'* direction are
reduced by a factor √(1-*β*^{2}) where *β=v*/*c*;
therefore *L'=L*√(1-*β*^{2}) so *t*_{1}'=2*L'*/*c*=2*L*√(1-*β*^{2})/*c=t*_{1}√(1-*β*^{2}).
For *L"* only its *x'* component is contracted, *L"*cos*θ'*=*L*√(1-*β*^{2})cos*θ
*while the *y"* component remains the same, *L"*sin*θ'*=*L*sin*θ*;
from these you can easily show that *L"=L*√(1-*β*^{2}cos^{2}*θ*)
and tan*θ'=*(tan*θ*)/√(1-*β*^{2}).
Finally, *t*_{2}*'*=2*L"*/*c*=[2*L*√(1-*β*^{2}cos^{2}*θ*)]/*c*=2*t*_{2}√(1-*β*^{2}cos^{2}*θ*)≠*t*_{1}*'*.

**ADDED
COMMENT: **I see that I misread your question. I see that
you have put the lengths at rest in the moving frame whereas I have them
in the stationary frame. But in special relativity it makes no
difference which is moving and which is at rest. So, you have found that
the times in the frame not at rest relative to the lengths are not the
same, just as I have; I reiterate that this is expected, not surprising.

**FOLLOWUP
QUESTION: **Thanks a lot, your competent and prompt
answer is much appreciated!

I would like to reiterate that for two events that occur ** in the same place** to be simultaneous in one frame but not in the other frame is really very strange, the non-existence of absolute simultaneity being a central tenet of Einstein's relativity not withstanding. For suppose that two light pulses when they arrive simultaneously to O' (according to a frame x'O'y' observer) they trigger an event, perhaps an explosion for which the simultaneous arrival of the two pulses is a necessary condition. So the explosion is the result of the simultaneous arrival of the two pulses for the x'O'y' observer but not for an observer in the frame xOy (who observes the explosion, too), since the latter sees the two pulses arriving at different times and thus cannot correlate the explosion to the two light pulses. The x'O'y' - observed concurrent emission of the two light pulses that go out from the very same point O' may be in an analogous way correlated to some event in the x'O'y' frame, but not in the xOy frame.

One can think with some justification that this scenario means that the x'O'y' system is somehow more "special" than the xOy system, pretty much in contradiction, if not with Einstein relativity postulate proper, at least with its spirit. In other words, in my opinion, the relativity theory is not devoid of inner contradictions, or at least it leads to conclusions which seem to invalidate the most fundamental notions of cause and effect or of things being what they are irrespective of where we observe them. And to me it is debatable whether a physical theory, however legitimately counter-intuitive it may be understood to be, can lead to this type of conclusions and be nevertheless considered a valid description of the physical world.

**ANSWER: **

Again, I would like to reiterate that it is neither strange nor
unexpected! First of all, the two events in the moving frame are at
neither the same time nor at the same place; only in the stationary
frame are the two events simultaneous and at the same place. By
"stationary frame" I mean here the frame where the apparatus is. Now,
your followup expands the apparatus to include also a "trigger" and a
bomb. The trigger will be some sort of electronic device to detect the
simultaneaty of the two light pulses' arrival at the origin and will
detonate the bomb in that case. An observer in the moving frame will not
see any contradiction, he will not say
that the bomb will not detonate because the device which detonates it is
in the stationary frame and he will agree that the pulses in that frame
are indeed simultaneous. Is the stationary system in some way "special"?
Of course it is—that is where the apparatus resides and all
frames will agree that the bomb will be triggered in that frame.

**QUESTION:***
*
An iron ball made of only iron ,no hollows no any other material,just iron. So first we measure its weight at room temperature and then we heat it up to higher temperatures near to melt but not not melting. Then we measure its weight. Does weights are different or same. If different please indicate which is higher.

**ANSWER: **

For purposes of calculation, let the ball have 1 kg of mass at room
temperature, 20°C. Because mass is a form of energy, the iron ball
has energy *E=mc*^{2}=9x10^{9} J. The melting
point of iron is 1540°C and its specific heat is 449 J/kg·°C,
so the enegy to heat it from 20 to 1540°C is 1x1520x449=6.8x10^{5}
J. So, the ball would have increased its mass by 6.8x10^{5}/9x10^{9}=7.6x10^{-5}
kg. You would be hard-pressed to actually measure so small a change in
mass.

*
***QUESTION:***
*
I have been trying to understand this for years. In Einstein's theory of time verses speed I believe he used the situation of a man (lets call him man A) standing in a train yard. A second man is on a train (lets call him man B) and the train is moving. They both drop an object at the same time. If I standing in the yard with man A to me the object man A drops would appear to me in normal time. Man B's object would appear to take longer causing the difference in time. I understand this. My main question is if a man C was on a different train going the same direction as man B's train the time difference between me and man A to man B and man C. I understand that to man B and Man c they would be the same. What if man C's train was going in the other direction? To me and man A the would seem the same, but man B to man C I believe the difference in the dropped object would be appear to be doubled to them between them. Since I believe it would appear doubled between man C and D. I think there should be a double time difference. From man A's view the age difference would be the same, but between man B and man C there should be double the time difference. How can there be doubled the time difference when they are traveling at the same speed?

**ANSWER: **

Your question boils down to what is called *velocity addition*.
In classical physics, *v*_{CB}=*v*_{CA}+*v*_{AB};
the notation is "*v*_{IJ} is the velocity of I relative
to to J". I have written this so that it corresponds to your question—A
is you, B and C are the trains; it is more convenient for you if we
rewrite the equation as
*v*_{CB}=*v*_{CA}*-v*_{BA}
which we can do because *v*_{AB}*=-v*_{BA}.
If the trains have speeds v in the same direction, *v*_{CB}=*v-v*=0;
in opposite directions, *v*_{CB}=*v-*(*-v*)=2*v—*each
sees the other moving with speed 2*v*. But, this form of velocity
addition is wrong for very high speeds (see an
earlier answer).
The relativistically correct velocity addition equation is
*v*_{CB}=(*v*_{CA}*-v*_{BA})/[(1-(*v*_{CA}*v*_{BA}/*c*^{2})]
which reduces to the classical equation for the speeds much less than
*c*. So, for the trains moving in opposite directions, *v*_{CB}=2*v*/(1+*v*^{2}/*c*^{2});
for example, if *v*=0.5*c*, *v*_{CB}=*c*/(1+0.25)=0.8*c*.

Now, you seem
to think that if you double the speed, you double the time difference.
This is not correct*—*time dilation goes like the gamma
factor, 1/√(1-*v*^{2}/*c*^{2}).
So, for your situation with *v*=0.5*c*, *t*_{A}=1.33*t*_{B},*
t*_{A}=1.33*t*_{C}, *t*_{C}=1.67*t*_{B},
and *t*_{B}=1.67*t*_{C}. These are
confusing, I admit. They are meant to denote, for example, that *t*_{B}=1.67*t*_{C
}means that B sees his clock tick out 1.67 s when C's clock ticks
out 1 s; B observes C's clock to be slow.

*
***QUESTION:***
*
While reading about the twin paradox, I've been told at the end of the traveling twin's journey, he begins decelerating in order to land back on Earth, and he, the traveling twin, observes his brother's clock on Earth to SPEED UP. This makes sense to me except for one problem: This suggests that the light pulse in the Earth clock would be percived by the traveling twin to be moving faster than C. Of course, the traveling twin is no longer in an inertial frame. I thought perhaps that since he feels himself moving now, he would also measure himself and the light having a CLOSING SPEED greater than C, even though he would see the light moving across the ground on Earth to equal C? If so, at what rate would a clock in frame S behind the traveling twin run at, faster or slower? Is it also possible that acceleration, from the point of view of the traveling twin, causes length contraction perpendicular to the the ship's vector, shortening the distance the pulse has to travel?

**ANSWER: **

There is no need to discuss acceleration to understand the
twin paradox. Acceleration
just makes everything harder to understand. Basically, just assume
necessary accelerations (departing, turning around, landing) occur in a
vanishingly short time. But, you are not really interested in the twin
paradox, you are interested in how things appear in an accelerated
frame. I am sorry, but nothing in your question after "…clock
on Earth to SPEED UP…" makes any sense. First of all, the fact
that any observer will measure the speed of light to be *c* is a
law of physics and the general principle of relativity states the laws
of physics are the same in *all* (not just inertial) frames.
Second, to measure a speed in your frame of reference you must use your
clock, not somebody else's. And third, how another clock looks is really
irrelevant because how it *appears* *to run* and how it
*is* *actually running* are not the same.

I would like
to address how the clocks look from the perspective of the
Doppler effect. The relativistically correct Doppler effect is
usually expressed in terms of the frequency of the light; for our
purposes, it is more convenient to express it in terms of the periods,
*T*_{observer}=*T*_{source}√[(1+*β*)/(1-*β*)]
where *β=v*/*c* and *β* is positive for
the source and observer moving apart; it makes no difference which is
the observer—each twin will see the other's clock running at the
same relative rate. Let's illustrate with a specific example, *β*=-0.8,
the traveling twin coming in at 80% the speed of light; *T*_{observer}=*T*_{source}√[(1-0.8)/(1+0.8)]=*T*_{source}/3
so the observer will see the source clock running fast by a factor of
three. But special relativity tells us that moving clocks run slow, not
fast; 1 second on the moving clock will be 1/√(1-0.8^{2})=1.67
seconds on the observer's clock. How the moving clock looks is thus
demonstrably not a measure of how fast it is actually running. Now, if
the incoming twin puts on the brakes such that he slows to *β*=-0.6,
*T*_{observer}=*T*_{source}√[(1-0.6)/(1+0.6)]=*T*_{source}/2
so the observer will see the source clock running fast by a factor of
two, apparently slowing down. Now, 1 second on the moving clock
will be 1/√(1-0.6^{2})=1.25 seconds on the observer's
clock, speeding up compared to when *β*=-0.8.

*
***QUESTION:***
*
If gravity is not a force but just a curvature of spacetime then how does a massive object (like earth) affect a much less massive object (like a tennis ball) when they are not in motion relative to each other?
For example, if I were able to travel out to space far enough away from earth and then stop so that I am not in motion with respect to earth and let go of a tennis ball it will stay stationary.
But if I traveled to say 100,000 ft above sea level and were able to hover there so I am not in motion to earth and then let go of a tennis ball it will immediately begin to move towards earth.
In other words, how is it possible for the curvature of spacetime to affect bodies that are not in motion to other?

**ANSWER: **

You have some misconceptions here. First of all, no matter how far away
you get, there will always be a small force toward the earth. It may be
so small that you would have to wait a millenium to see it move a
millimeter, but it is still there. The curvature of spacetime justs gets
smaller as you get farther away. It is certainly true that if you drop
something from an altitude of 100,000 feet it will accelerate toward the
earth. But, what does that have to do with motion? Even if you gave it
some motion, say throwing it horizonatlly, it would still accelerate
toward the earth just the same as dropping it, but now it would also
have a velocity component parallel to the earth as well. Maybe you have
never seen the "trampoline
model" which illustrates (in a simplistic, not literal way) how
warping the space (the surface of a trampoline) by a massive object
(bowling ball) will cause a light object (marble) to be attracted to the
massive object.

*
***QUESTION:***
*
So if black holes suck in everything in including light that must mean everything is getting pulled in as fast or greater then the speed of light. So if light is weightless and it is sucked in. What happens to any mass as it is sucked in. Would the mass of the object then cease to have mass? Because im pretty sure anything traveling at the speed of light has to be mass-less correct? And how does gravity effect something with no mass? I dunno if it is a good question or not but i couldnt find a whole lot on the subject.

**ANSWER:
**

Nothing ever goes faster than the speed of light and only light
can go at the speed of light. When an object merges with the black hole,
its energy, *E=mc*^{2}, is not lost and the black hole
becomes more massive by the amount of the mass energy. When light is
swallowed by the black hole, the mass increases even though the light
does not have any mass because light does have energy and the energy
shows up as increased mass of the black hole, *m=E*/*c*^{2}.

*
***QUESTION:***
*
When speaking of particle accelerators,the accelerators keep adding energy to the particles, even though they cannot speed up any further. But where does the energy go?

**ANSWER:
**

Well, they never really get to a speed where they cannot go any
faster because they never reach the speed of light. I have always
thought "accelerator" was a misnomer for very high-energy machines
because the acceleration (rate of change in speed) is almost zero; I
would prefer "energizer". The easiest way to think about it is that the
particles acquire mass and that is "…where…the energy
go[es]". The energy of the particles is *E=mc*^{2}=*m*_{0}*c*^{2}/√[1-(*v*/*c*)^{2}]
where *m*_{0} is the mass at rest. For example, if *v*/*c*
increases from 0.99 to 0.999, that is only a 1% change in speed; but the
energy increases from 7.1*m*_{0}*c*^{2}
to 22.4*m*_{0}*c*^{2}, more than triple.

*
***QUESTION:***
*
Does mass of an object increase its mass exponentially as it approaches infinitely close to the speed of light?

*
***ANSWER:
**

If d*f*/d*q*=*Aq* for some function *f*(*q*),
where *A *is a positive constant, *f* is said to be
exponentially increasing with *q*. For your question, *f*
is *m* and *q* is *v*. The expression for *m*(*v*)
is* m=m*_{0}/√[1-*v*^{2}/*c*^{2}]
where *m*_{0} is the rest mass. Calculating the
derivative, d*m*/d*v*=*mv*/[*c*^{2}-*v*^{2}].
Thus, although *m* increases without bound as *v*
increases, it does not increase exponentially.

**QUESTION:**

A starship pilot wants to set her spaceship to light speed but the crew and passengers can only endure a force up to 1.2 times their weight. Assuming the pilot can maintain a constant rate of acceleration, what is the minimum time she can safely achieve light speed?

*
***ANSWER:**

This question completely ignores special relativity. It is impossible to go
as fast as the speed of light. Furthermore, acceleration is not really a
useful quantity in special relativity and you must use special relativity
when speeds become comparable to the speed of light. I have
earlier worked out the
velocity of something which would correspond to occupants of your spaceship
experiencing a force equal to their own weight due to the acceleration which
I will adapt to your case later. (See the graph above.) First, though, I
will work out the (incorrect) Newtonian calculation which is presumably what
you want. The appropriate equation would be *v=at* where *v*=3x10^{8}
m/s, *a*=1.2*g*=11.8 m/s^{2}, and *t* is the time to
reach *v*; the solution is 2.5x10^{7} s=0.79 years. For the
correct calculation, you cannot reach the speed of light; from the graph
above (black curve), though, you can see that you would reach more than 99%
of* c* when*
gt*/*c*=3.
To make this your problem, we simply replace *g* by 1.2*g* and
solve for *t*. I find that *t*=7.7x10^{7} s=2.4 years.

**QUESTION:**

I am a high school student interested in relativity and I recently read an
article
about relativity. The article stated that the "The combined speed of any object's motion through space and its motion through time is always precisely equal to the speed of light." For an object moving at 90% the speed of light, it should only be moving at 10% the speed of light through time. I assume that means that time should pass only at 10% the actual speed of time for the object (correct me if I am wrong). However based on the Lorenz factor, time passes at 0.43 the actual speed of time for the object. Why is this so?
Furthermore when time dilation occurs it is only seen by someone else in a stationary frame of reference. In the moving object's frame of reference time does not slow down at all. Does this mean that the combined speed of the moving object's motion through space and time can be more than the speed of light in the moving object's frame of reference? Can the moving object go beyond the speed limit in its frame of reference?

**ANSWER:**

Look closely at the example given: "*To get a fuller sense of what Einstein found, imagine that Bart has a skateboard with a maximum speed of 65 miles per hour. If he heads due north at top speed—reading, whistling, yawning, and occasionally glancing at the road—and then merges onto a highway pointing in a northeasterly direction, his speed in the northward direction will be less than 65 miles per hour. The reason is clear. Initially, all his speed was devoted to northward motion, but when he shifted direction some of that speed was diverted into eastward motion, leaving a little less for heading north.*"
The speed in the northward direction will now be *v*_{N}=65cos45^{0}=46
mph; his velocity in the eastward direction will be *v*_{E}=65sin45^{0}=46
mph. His total velocity will be √(*v*_{N}^{2}+*v*_{E}^{2})=65
mph, not (*v*_{N}+*v*_{E})=92
mph. In relativity N and E now are *x* and *t* for the
stationary observer and *x'* and *t' *for the position and
time of the moving observer as seen by the stationary observer. The catch,
though, (which I presume is why the author of your article avoided these
details) is that time and space are slightly different from space (N) and
space (E) in that to get the length of the vector you calculate the square
root of the difference of the squares instead of the sum. So the speed
through spacetime would be √(c^{2}-v^{2}).
So, the stationary observer will see a "spacetime speed" of *v*_{spacetime}=√(c^{2-}v^{2})
and the "space time distance traveled" by the moving observer as seen by the
stationary observer would be *d*_{spacetime}=*v*_{spacetime}*t'*
where* t'=t*/√(1-(*v*^{2}/*c*^{2})).
If you do your algebra you will see that *d*_{spacetime}=*ct
*indicating that the stationary observer sees the moving observer having
a speed of *c* through spacetime.

**QUESTION:**

When a photon loses energy ( Redshifts) climbing up thru a gravitational field, does the photons decreased energy go into the mass of the gravitational field itself? According to GR, the gravitational field itself contributes to the mass of the system. For earth, it is very small. When I elevate a massive object in a gravitational field, I can say the "potential energy" is in a very small increase of the mass of the object. But photons have no mass! And are never at rest! The only thing I can think of is that the photon "transfers" energy to the gravitational field itself, which appears as a small increase in mass of the field.

**ANSWER:**

Let's think of a star emitting a photon of frequency *f*. Initially
an energy of *hf* is removed from the star. But, by the time that
the photon is very far away the star, it has lost some amount of energy Δ*hf*,
so the net loss to the star and its field is
*hf*-Δ*hf*. At the instant the photon is created, the
mass of the star is reduced by *hf/c*^{2}. But when the
photon, which is losing its energy to the field, is far away, the *energy*
of the field will have increased by Δ*hf*/*c*^{2};
I would not use the terminology "mass of the field" since the field has
energy density, not mass. But now, it seems to me (not a cosmology/general
relativity expert) that there is a bit of a paradox: we always associate a
given mass with a particular gravitational field, so the field should have
the energy content associated with a mass *M'=M*-*hf/c*^{2}.
But, in fact, the field would have energy content associated with a mass *
M'=M*-*hf/c*^{2}+Δ*hf*/*c*^{2}.
I am guessing that the field and the mass somehow "equilibrate" so that the
final mass of the star is consistent with the energy of the final field. (I
could easily be wrong! Perhaps it is only meaningful to look at the total
energy of the star and its field. Whatever the case, I would not talk about
mass of the field.)

**QUESTION:**

Sir,I am a mechanical engineer by profession but very interested in reading physics fundamentals. Recently I went through the fundamentals of electro magnetism and I got this doubt. Consider two charges each of charge +q rigidly fixed in a train moving with a constant velocity, V. Let the train speed be negligible compared to speed of light so that we can treat the problem in non-relativistic terms. The fixity condition ensures that it overcomes electrostatic forces and remain motionless. A traveler in train sees both of them at rest and there wont be any magnetic forces developed between the charges.
Now consider an observer in platform. For him, both charges are moving with a constant velocity, V equal to the train velocity. Each charge will develop magnetic field according to this observer as per Biot-Savart law and there will be mutual attraction as each charge is moving under the magnetic field of other.Thus, an observer in train sees no magnetic forces whereas an observer in platform sees mutual magnetic attraction. How do you explain this?

*
***ANSWER:**

There is no rule which says that the either the electric or magnetic field
must be the same in all frames of reference, even slowly moving frames like
you specify. The real root of your problem is that electromagnetism is
intrinisically relativistic, even at slow speeds; the electric and magnetic
fields of classical electromagnetism are really both components of the
electromagnetic field which is a tensor and when you change inertial frames,
you cause a transformation of that tensor into another where both the
electric and magnetic field pieces of it are different. In your second case
you would also find that the electric fields were slightly different from
their original values but the differences would be very tiny; the magnetic
fields, though, are nonzero but small, but small is very big compared to the
original magnetic field of zero.

If you
are interested, I will give here the electric and magnetic fields for one of
the charges moving with velocity * v* in the +

*x*direction.

**=**

*E'*

**i**E_{x}+

*γ*(

**j**E_{y}+

**k**E_{z}) and

*(*

**B'**=-*x*

**v***)/*

**E'***c*

^{2 }where

*γ=*1/√[1-(

*v*/c)

^{2}] and

**,**

*i***,**

*j***are unit vectors; the vector**

*k**is in the frame where*

**E***q*is at rest and

**and**

*E'**are when*

**B'***q*is moving.

**QUESTION:**

Is it true that a person flying in an airplane is actually living a shorter amount of time, than a person standing on the ground?

*
***ANSWER:**

You must specify "…living
a shorter amount of time…"
with respect to whom . The
person in the airplane sees time progressing at a normal rate. However, the
person on the ground will see the clock of the person on the airplane run
slowly, so he will perceive the traveling person to be " living
a shorter amount of time", *i.e.* she will have aged less when she
returns to earth. You should read FAQ Q&As on the
twin paradox,
how clocks run, and the
light clock.

**QUESTION:**

I think if an object is turning, it has more gravity than an object which is not turning

*
***ANSWER:**

You are right. However, it really has nothing to do with the turning, *per
se*. When something is turning it has rotational kinetic energy and
therefore a spinning planet has more energy than an otherwise identical
planet not spinning. In general relativity you usually hear about gravity
being caused by mass warping spacetime. However, mass is just the most
obvious source of gravity and what it is which really warps spacetime
is energy density and mass has a lot of energy (*E=mc*^{2}). The energy due to the turning
is infinitesmal compared to the mass energy of the object so you would never
be able to distinguish the difference due to the turning by looking at the
gravity.