- #1
cen2y
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I do not here state that srt or it's applications would be wrong but simply I wish to bring out a faulty example.
In the text box I carried about a year ago, the following "thought-experiment" and example were brought to support the mass-energy equivalence:
--- Einstein's gedanken, the lamp in the box example, as written by me. May be skipped if you know it. ---
Consider having a box B of the length L, carrying a lamp L. If the lamp at a given point would fire a photon along the box, the photon would carry the linear momentum p=E/c. In order to proserve the tot. linear momentum, the box must experience a velocity of v=p/M=E/(Mc). The photon taking the time t=L/(c(1+v/c)) to reach the far end, delivering E while moving the box the distance d = tv = Et/(Mc) = EL/(Mc²(1+v/c)), in order for the box's com to be static, the mass of the box must've rearranged. Given that the box is symmetrical and the only change would be the movement of the photon, a mass m must've traveled from the position -L/2 to L/2, causing the com to move the distance d = EL/(Mc²(1+v/c)), thus EL/(Mc²(1+v/c)) = mL/(M(1+v/c)) => m = E/c² (mL/(M(1+v/c)) being the movement of the com given the mass m is transferred from one end to the other).
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However, one must account that the center of gravity did not but move until the striking of the photon, but continuosly during it's travel. One may therefor deduct that
dcom(t) = tv = Et/(Mc)
as it is but the photon moving, one will find that Et/(Mc) = mct/M => m = E/c², where m is the systematic mass of the photon. This of course doesn't mean that a photon would have all types of mass, but nevertheless it either has a systematic mass and/or the formula for com ought to include energy as well as this example ought not to be enough to "deduct" the mass-enegy equivalence.
Edit: Forgot to take into account the box's movement as the photon travels and wrote cog instead of com.
In the text box I carried about a year ago, the following "thought-experiment" and example were brought to support the mass-energy equivalence:
--- Einstein's gedanken, the lamp in the box example, as written by me. May be skipped if you know it. ---
Consider having a box B of the length L, carrying a lamp L. If the lamp at a given point would fire a photon along the box, the photon would carry the linear momentum p=E/c. In order to proserve the tot. linear momentum, the box must experience a velocity of v=p/M=E/(Mc). The photon taking the time t=L/(c(1+v/c)) to reach the far end, delivering E while moving the box the distance d = tv = Et/(Mc) = EL/(Mc²(1+v/c)), in order for the box's com to be static, the mass of the box must've rearranged. Given that the box is symmetrical and the only change would be the movement of the photon, a mass m must've traveled from the position -L/2 to L/2, causing the com to move the distance d = EL/(Mc²(1+v/c)), thus EL/(Mc²(1+v/c)) = mL/(M(1+v/c)) => m = E/c² (mL/(M(1+v/c)) being the movement of the com given the mass m is transferred from one end to the other).
------
However, one must account that the center of gravity did not but move until the striking of the photon, but continuosly during it's travel. One may therefor deduct that
dcom(t) = tv = Et/(Mc)
as it is but the photon moving, one will find that Et/(Mc) = mct/M => m = E/c², where m is the systematic mass of the photon. This of course doesn't mean that a photon would have all types of mass, but nevertheless it either has a systematic mass and/or the formula for com ought to include energy as well as this example ought not to be enough to "deduct" the mass-enegy equivalence.
Edit: Forgot to take into account the box's movement as the photon travels and wrote cog instead of com.
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