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Why can't an object moving at the speed of light be taken as reference frame?

Can we use the equation m=m(0)/sqrt(1-v^2/c^2) for an object moving with speed of light?

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Why can't an object moving at the speed of light be taken as reference frame?

Can we use the equation m=m(0)/sqrt(1-v^2/c^2) for an object moving with speed of light?

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Part of the answer directly follows from your questions: using your equation, you will find that only an object with zero rest mass can propagate at the speed of light; and such objects are called photons (it is assumed that photons have exactly zero rest mass). Note that as 0/0 is useless, for the "mass equivalent" of light you can use m=p/c.

Why can't an object moving at the speed of light be taken as reference frame?

Can we use the equation m=m(0)/sqrt(1-v^2/c^2) for an object moving with speed of light?

And how would you use a photon as reference frame? A reference frame is a system for comparing (measuring) such things as time and distance. If a clock and ruler would be accelerated to light speed (although impossible), they would stop ticking and have zero length.

- #3

Dale

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here is a FAQ which explains why.Why can't an object moving at the speed of light be taken as reference frame?

https://www.physicsforums.com/showthread.php?t=511170

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- #5

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Again, use your own equation! How much relativistic mass will it have at the speed of light? How much energy is needed to bring it to that speed?

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Yes, other massless bosons.Is there any other object except photon which moves at the speed of light?

Your question has been answered quite well here. Also, you might consider the problem in the context of space-time diagrams (google it or find discussions of space-time diagrams in other posts). The sketches below show a sequence in which an observer (blue frames of reference) moves at ever greater relativistic velocities with respect to a rest frame (black perpendicular coordinates). One aspect of the photon (any massless boson) that makes it so special is that its worldline always bisects the angle between the time axis and the spatial axis for any observer, no matter what the observer's speed (thus, the speed of light is the same for all observers).Why can't an object moving at the speed of light be taken as reference frame?

Notice in the sequence that the moving observer's X4 and X1 axes rotate toward each other, getting closer and closer to each other as the speed of light is approached. In the limit the X4 axis and the X1 axis overlay each other. So, if the observer were actually moving at the speed of light, both his time axis and his spatial axis would be colinear with the photon worldline. How would you define that as a coordinate system?

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I am not talking of an observer moving at the speed of light rather I am talking about an observer observing an object x w.r.t to an object y(moving at the speed of light).Yes, other massless bosons.

Your question has been answered quite well here. Also, you might consider the problem in the context of space-time diagrams (google it or find discussions of space-time diagrams in other posts). The sketches below show a sequence in which an observer (blue frames of reference) moves at ever greater relativistic velocities with respect to a rest frame (black perpendicular coordinates). One aspect of the photon (any massless boson) that makes it so special is that its worldline always bisects the angle between the time axis and the spatial axis for any observer, no matter what the observer's speed (thus, the speed of light is the same for all observers).

Notice in the sequence that the moving observer's X4 and X1 axes rotate toward each other, getting closer and closer to each other as the speed of light is approached. In the limit the X4 axis and the X1 axes overlay each other. So, if the observer were actually moving at the speed of light, both his time axis and his spatial axis would be colinear with the photon worldline. How would you define that as a coordinate system?

I didn't get well this graphical explanation, would you please elaborate.

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HallsofIvy

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Then you will have to explain what you mean by that. What do you mean by "observing x with respect to y"? Any observer see object with respect to himself, not with respect to any other frame of reference.I am not talking of an observer moving at the speed of light rather I am talking about an observer observing an object x w.r.t to an object y(moving at the speed of light).

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I don't think that is necessary, let's talk as an example about 3-Dimensinal co-ordinate system in which an observer observes the motion of any object w.r.t to the origin(0,0,0).Then you will have to explain what you mean by that. What do you mean by "observing x with respect to y"? Any observer see object with respect to himself, not with respect to any other frame of reference.

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Nugatory

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Ah, but is the observer moving with respect to that origin? If so, we have three frames (observer, object, and frame-with-origin-at-(0,0,0)) to transform between, not two. None of these frame can have a velocity greater than or equal to to the speed of light relative to any other of these frames.I don't think that is necessary, let's talk as an example about 3-Dimensinal co-ordinate system in which an observer observes the motion of any object w.r.t to the origin(0,0,0).

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The observer is at rest w.r.t origin.Ah, but is the observer moving with respect to that origin? If so, we have three frames (observer, object, and frame-with-origin-at-(0,0,0)) to transform between, not two. None of these frame can have a velocity greater than or equal to to the speed of light relative to any other of these frames.

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Dale

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Yes, any object with non zero rest mass must move slower than c in any inertial frame.

As far as why, that is inherently a tricky question. What are you allowing to be assumed when answering? And what kind of answer are you looking for? If I were asking the question I would be looking for a geometric answer and I would allow the Minkowski metric to be assumed. Then the answer is that a massive object has a timelike four momentum by definition, and any timelike four momentum corresponds to a three velocity < c. If that doesn't answer the question then you will need to clarify what you want better.

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OK, then my answer of post #5 applies. How much energy do you think is required?I don't think that is necessary, let's talk as an example about 3-Dimensinal co-ordinate system in which an observer observes the motion of any object w.r.t to the origin(0,0,0).

The observer is at rest w.r.t origin.

If you did not manage to calculate that a division by zero is infinite, the answer is given in section 10 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ :

Thus, when v=c, [kinetic energy] W becomes infinite

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