# Insufficient example, einstein's gedanken: lamp in the box

1. Jan 24, 2005

### cen2y

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 transfered 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.

Last edited: Jan 25, 2005
2. Jan 24, 2005

### K.J.Healey

You say it moves the box a distance of d = t*v but what is t here? I assume it would be the duration of time for which the box moves. Why would this be in anyway equivalent to the time it takes for the photon to travel from the source to the box(L/c)? That seems like an error due to notation.

3. Jan 25, 2005

### cen2y

The reason it'd be the same amount of time as it takes for the photon to reach the far end of the box would be as the linear momentum of the photon would then be absorbed, forcing the box to a halt. Though L/c would be an incorrect time of travel, a miss by me, I'll update the post. Nevertheless if that formula was misinterpreted or not, it would not explain as though why the com would move while the photon travels.

4. Jan 25, 2005

### K.J.Healey

I think im missing the general issue of the problem...
Can we not resort to imagining ourselves floating in the box, and throwing a ball toward the wall? Assuming we're not attached to the box, why does the CoM of the box have to change at all until the impact?

5. Jan 26, 2005

### Ivan Seeking

Staff Emeritus
Note that this assumes a perfectly rigid box.

6. Jan 26, 2005

### cen2y

The com, seen from an external viewpoint, will be dislocated from it's original position as the box must travel due to it's increament of velocity, caused by the increment in linear momentum, caused by the projection of light. As no internal process may alter the com of an object, mass must be rearranging relative the com during the movement of the box.

If you mean per shape, except for the mass of transfer, it assumes a rigid box, yes.
Though if one would mean rigid as in a box fixed to a certain position, rather the opposite would cause a problem, as though only one is incapable to speak of the box's movements. Where the position is measured in relation to a body not fixed relative the system of reference however, one may simply account the box's mass as the complete body, given no angular momentum would be required. Though neither of this is of any interest to the initial inquire.
On a sidenote, you do carry quite some quotes there.

7. Feb 7, 2005

### cen2y

I forgot to replace v in the equation above, here's the complete equation:

If we call
$v$ the speed of the box,
$M$ it's mass,
$L$ it's length, from one end to the other,
$p_{\textit{box}}$ it's linear momentum,
$d$ the distance the box is dislocated,
$\Delta cm_{\textit{x,box}}$ the dislocation of it's center of mass due to the box's movement,
$\Delta cm_{\textit{x,light}}$ the dislocation of it's center of mass due to the light's movement,
$t$ the time it, as well as the light, moves,
$E$ the energy of the lightbeem/photon,
$m$ the photon's transfer of mass and
$p_{\textit{box}}$, we recieve the following equations:
$$\begin{array}{l} \left\{ \begin{array}{l} \Delta p_{\textit{light}}+\Delta p_{\textit{box}}=0 \\ \Delta p_{\textit{light}} = E/c \\ \Delta p_{\textit{box}} = Mv \end{array} \right \Rightarrow v = -\frac{E}{cM} \\ \left\{ \begin{array}{l} d = L + vt = L - \frac{Et}{cM} \\ t = d / c \end{array}\right \Rightarrow t = \frac{L-\frac{Et}{cM}}{c} = \frac{L}{c}-\frac{Et}{c^{2}M} \Leftrightarrow t(1+\frac{E}{c^{2}M})=\frac{L}{c} \Leftrightarrow t = \frac{L}{c(1+\frac{E}{c^{2}M})} = \frac{cL}{c^{2}+\frac{E}{M}} \\ \left\{ \begin{array}{l} \Delta cm_{\textit{x,box}}+\Delta cm_{\textit{x,light}}=0 \\ \Delta cm_{\textit{x,box}}=vt=-\frac{E}{cM}\frac{cL}{c^{2}+\frac{E}{M}}=-\frac{EL}{c^{2}M+E}=-\frac{L}{1+\frac{c^{2}M}{E}} \\ \Delta cm_{\textit{x,light}}=\frac{dm}{M}=\frac{m(L-\frac{Et}{cM})}{M}=\frac{m(L-\frac{cEL}{cM(c^{2}+E/M)})}{M}=\frac{m(L-\frac{L}{1+c^{2}M/E})}{M}=\frac{mL}{M}(1-\frac{1}{1+c^{2}M/E}) \end{array} \Rightarrow \\ \Rightarrow -\frac{L}{1+c^{2}M/E} = \frac{mL}{M}(1-\frac{1}{1+c^{2}M/E}) \Leftrightarrow M=m(1+c^{2}M/E)(1-\frac{1}{1+c^{2}M/E})= \\=m(1+c^{2}M/E-1)=c^{2}mM/E \Leftrightarrow m=\frac{EM}{c^{2}M}=\frac{E}{c^{2}} \end{array}$$

Due to continous movement however:
$$m_{\textit{light}}(t)=E/c^{2}$$

Last edited: Feb 7, 2005