# Light speed laser pointer war

1. May 27, 2009

### gonegahgah

I feel kinda silly asking this question.

If you were traveling near the speed of light towards another ship and you shone one of those simple red lasers towards that ship would that ship blow up?

2. May 27, 2009

### sylas

If you were going fast enough, the laser would be blue shifted to hard gamma radiation. It might not blow up, but you could fry it pretty severely. Also, the intensity of the radiation received by the enemy would be enormously increased (photons per second) as long as you were able to keep the laser on for long enough. (That is, all the photons you emit within a second would be received by the enemy within a much shorter period of time.)

Cheers -- sylas

3. May 27, 2009

### gonegahgah

Thanks Sylas

4. May 27, 2009

Staff Emeritus
However, you would probably have to leave the laser pointer pointed at your target for a very long time: years or centuries. Energy is still conserved, and these lasers only put out a fraction of a watt.

5. May 27, 2009

### sylas

Energy depends on the frame of reference. Each photon has more energy from the perspective of the enemy than from the perspective of the laser. You get the conversion into gamma radiation for free. Both observers agree on the number of photons; though not on the time it takes to emit them, or absorb them.

The total energy boost scales as the redshift, which is
$$\sqrt{\frac{c+v}{c-v}}$$
Suppose you have a 5 mW laser, and you shine it for 200 seconds on the enemy. That's a Joule. If you want that to deliver the energy of a good sized nuke (1015 J or so), then you want to be traveling at about 2*10-30 short of the speed of light. That is,
$$v/c = 1 - 2*10^{-30}$$

Cheers -- sylas

PS. BTW... at that speed, you've got a few gamma radiation problems of your own. Cosmic background radiation has a wavelength of around 2mm, or 6.2*10-4 eV per photon. You'll be running into that radiation yourself... and it is now blueshifted by a factor of 1015 to be 2*10-18 m, or over 600 GeV per photon. That's about as hard as gamma radiation gets! The intensity will also be enormous, thanks to your velocity.

Last edited: May 27, 2009
6. May 27, 2009

### Count Iblis

7. May 27, 2009

### Bob S

Here is an analagous situation. Physicists at SLAC (Stanford Linear Accelerator Center) aim a pulsed visible laser at an electron beam comming toward them with an energy = 40 GeV = gamma of 80,000. The photons bounce off the oncomming electrons (Compton scattered) and come back to the experimenters, boosted in energy by a factor of gamma squared (~6 GeV). Beware of alien spaceships with photon shields (Compton reflectors).

Last edited: May 27, 2009
8. May 27, 2009

### Count Iblis

If the alien spaceships are accelerating then the reflectors will emit photons due to the dynamical Casimir effect.

9. May 28, 2009

Staff Emeritus
Yes, but energy is still conserved.

If body A transfers a joule to body B, body B receives one joule. The energy transferred is in fact an invariant quantity, even if kinetic energy itself is frame dependent.

So where does the energy come from? You're assuming body B is at constant velocity, but the laser transfers momentum as well as energy. To keep the velocity constant, you also have to fire the engines: and that's where the energy comes from.

10. May 28, 2009

### Bob_for_short

I see the first time a Vanadium's error. Normally his posts are wise and knowledgeable.

The energy is conserved in time in a certain reference frame. The transferred energy is a frame depended quantity. The energy transferred by laser to photons (red light seen in a fast ship) is not the energy received by an enemy target (hard gamma rays). Forget about photons. The same is valid for any projectile. Even though you through a piece of cake with your hand with a small velocity, it will hit the enemy hard. The hit energy is a part of your ship initial energy, and the enemy always sees it as large (it is a kinetic energy of the relative motion).

You in your ship experience the same thing: the enemy flies rapidly towards you, it has already a great deal of kinetic energy. It suffices to through a piece of cake towards it to destroy its ship.

Bob_for_short.

Last edited: May 28, 2009
11. May 28, 2009

### neopolitan

Time dilation must be a factor. Your laser is pumping out energy in watts, 5W is a pretty powerful one. That is 5 joules per second.

You are in your ship travelling at 0.8c, relative to your enemy. That means you are time dilated by 0.6 so you are putting out 3 joules per second, relative to your enemy.

There's a doppler effect as well, though. Without checking, I am going to make a wild stab and say you have an increased frequency by 1/0.6 which means the associated energy per photon is increased by the same, so the wattage received by your enemy is 3/0.6 = 5 joules per second.

I may be wrong on that, but gut feeling tells me Vanadium had it right.

cheers,

neopolitan

12. May 28, 2009

### Bob_for_short

No, the time dilation affects the power = energy/time. It will increase the received energy!

And power (&E/&t) is not a relativistic invariant.

The energy h_bar*omega (or E(V_relative), whatever) is frame dependent. Namely due to the Doppler effect for photons. It will also increase the received energy!

Bob_for_short.

Last edited: May 28, 2009
13. May 28, 2009

### neopolitan

I don't think so.

Each photon has more energy in your enemy's frame, but you are pumping out photons more slowly in your enemy's frame because your time is dilated in that frame. The overall effect is that the wattage is invariant.

I think ...

cheers,

neopolitan

14. May 28, 2009

### sylas

No, the energy transferred from the laser to the target is not invariant. It is -- like the kinetic energy -- frame dependent. Energy is conserved in all frames, but the frames differ in the transfer amount of energy involved.

The actual invariant is the new rest mass of the target after it has absorbed the laser blast. That's the measure of actual damage done. But where you balance that energy differs in different frames! From the perspective of the target, it gets a whole heap of energy from the incoming gamma radiation, which packs a heck of a punch from the blueshift. Nearly all the energy here is delivered from the laser.

But from the view of the laser, it only sent out a weak little photon, at very little energy cost to itself. Almost all the energy for "blowing up" the target is actually taken from the target's own kinetic energy as it ploughs into those photons. From the point of view of the laser, it just left a few weak little photons in the way of the target, which smashed into them with high velocity!

No, I am not presuming constant velocity.

Cheers -- sylas

Last edited: May 28, 2009
15. May 28, 2009

### sylas

OK guys... here's an exact calculation.

A laser, and a target, each have rest mass M. They approach each other at velocity v, with a gamma factor γ. From the perspective of either laser or target, the other one is moving, and has a total energy of γ.M.c2. The momentum p of the moving mass satisfies the following:
$$\begin{equation*}\begin{split} (\gamma M c^2)^2 & = (pc)^2 + (M c^2)^2 \\ (\gamma^2 - 1) (M c^2)^2 & = (pc )^2 \\ \frac{(v/c)^2}{1-(v/c)^2} (M c^2)^2 & = (pc )^2 \\ \gamma M v & = p \end{split}\end{equation*}$$​

In general, for a particle of energy E and momentum p, you can find the rest mass as
$$m = \sqrt{(E/c^2)^2 - (p/c)^2}$$​

Now the laser fires a single photon at the target, with frequency f in the frame where the laser is initially at rest. The laser recoils, and has a reduced rest mass. The photon goes on to collide with the target, which is slowed by the impact, but gains rest mass.

Case 1. Analysis in the frame where the laser is initially at rest

The momentum of the photon is hf/c, and the energy of the photon is hf. By conservation of momentum, the laser recoils, with momentum -hf/c. It also has a total energy of M.c2-hf, in this frame.

The new rest mass of the laser is
$$\begin{equation*}\begin{split} m_l & = \sqrt{(M - hfc^{-2})^2 - (hfc^{-2})^2} \\ &= \sqrt{M^2 - 2Mhfc^{-2}} \\ & \approx M - hfc^{-2} \end{split}\end{equation*}$$​
I've given the exact new rest mass; but as you can see it is about what you expect to give up the energy required to fire the photon. The energy of the recoiling laser is negligible by comparison.

Meanwhile the photon collides with the target. The target now has momentum of magnitude γ.M.v - hf/c, and energy of γ.M.c2 + hf.

(Note that v and γ are treated as constant throughout, reflecting the velocity before the photon exchange!)

Hence the new rest mass of the target is
$$\begin{equation*}\begin{split} m_t & = \sqrt{(\gamma M + hfc^{-2})^2 - (\gamma M v/c - hfc^{-2})^2} \\ &= \sqrt{(\gamma^2 - (v/c)^2 \gamma^2 )M^2 + 2(1+v/c)\gamma Mhfc^{-2}} \\ &= \sqrt{M^2 + 2(1+v/c)\gamma Mhfc^{-2}} \\ & \approx M + (1+v/c)\gamma hfc^{-2} \end{split}\end{equation*}$$​
The factor (1+v/c)γ turns out to be precisely the factor of the Doppler blue shift. That is, the target has gained in rest mass as if colliding with a blueshifted photon. In the frame of the laser, most of this energy comes NOT from the laser, or from the any firing of engines, but from kinetic energy lost by the target as it ploughs into the photon!

Case 2. In the frame where the target is initially at rest

The photon is blueshifted, by a factor
$$\sqrt{\frac{c+v}{c-v}} = (1+v/c)\gamma$$​
The momentum of the photon is (1+v/c)γhf/c, and the energy of the photon is (1+v/c)γhf. By conservation of momentum, the laser recoils, and now has momentum
$$p = \gamma M v - (1+v/c) \gamma hf/c$$​
The new energy of the laser, in this frame is
$$E = \gamma M c^2 - (1+v/c) \gamma hf$$​
Hence the new rest mass of the laser after delivering this very energetic photon is
$$\begin{equation*}\begin{split} m_l & = \sqrt{(\gamma M - (1+v/c)\gamma hfc^{-2})^2 - (\gamma M v/c - (1+v/c) \gamma hfc^{-2})^2} \\ &= \sqrt{\gamma^2(1-(v/c)^2)M^2 - 2\gamma^2M(1+v/c)hfc^{-2}(1-v/c)} \\ &= \sqrt{ M^2 - 2 M hfc^{-2}} \\ & \approx M - hfc^{-2} \end{split}\end{equation*}$$​
... as before. Most of the energy in the incoming photon is actually taken up from energy lost by the recoil of the laser, in this frame!

The new rest mass of the target, from the energetic photon, is
$$\begin{equation*}\begin{split} m_t &= \sqrt{ ( M + (1+v/c) \gamma hfc^{-2})^2 - ( (1+v/c) \gamma hfc^{-2})^2} \\ &= \sqrt{ M^2 + 2 \gamma M (1+v/c) hfc^{-2} } \\ &\approx M + \gamma (1+v/c) hfc^{-2} \end{split}\end{equation*}$$​
... as before.

Cheers -- sylas

PS. Once you understand what happens for a single photon, generalizing is easy. The total number of photons delivered is invariant. Hence the total energy is scaled by the factor
$$\sqrt{\frac{c+v}{c-v}}$$​
The time dilation means that this energy is also received over a smaller interval of time than it took to deliver it.

For example. You fire your 5mW laser for one minute, at a target approaching at 80% light speed. The blueshift is a factor of 3. You are expending 0.3 Joule, to deliver 0.9 Joules of damage.

The target gets energy at 45 mW, over a span of 20 seconds. It is receiving photons three times as rapidly, for one third as long, but each one packing three times the punch.

Last edited: May 28, 2009
16. May 28, 2009

Staff Emeritus
Let me further add to the confusion (I tried to use some shorthand and skip some steps but it might not have been such a good idea in retrospect):

First let me clarify, the thing that's actually invariant is Mandelstam t, which is not exactly the kinetic energy transferred, but it is in the limit.

If I am in a single frame, and body A gives energy to body B, the energy emitted by A has to equal the energy absorbed by B. Does anyone doubt this?

So if body A tosses a joule's worth of photons at body B, in that frame, B can only gain a joule of energy. When we see more energy, it had to come from somewhere. But where? The answer is, B's engines. These photons not only carry energy, they carry momentum. So B slows down - the energy that goes into the destruction did not come from the laser pointer, but instead came from B's engines. If you let B slow down, it's energy the engines expended in the past. Alternatively, you can keep B's speed constant if the engine's thrust makes up the difference.

For a given power for a laser, the energy transferred depends on two things: the time it's on and the speed of the target. You have to either leave the laser on for a long time (at gamma of 60, a minute of time at the laser pointer end becomes only a second of time as observed at the targete end) or have the target go even faster.

17. May 28, 2009

### sylas

THREE things. You've omitted the intensity of the beam, which depends on the frequency of the photons and on the number per unit time.

The energy "transferred" is DIFFERENT depending on who is measuring it. This was the error in your initial post, when you said that the transferred energy was the invariant. It isn't.

Try again. You have a 5 milliWatt laser, that you shine for one minute at an incoming target moving at 80% light speed.

OK? To make SURE you don't tie yourself up in knots, calculate the energy you've transferred. It is 0.3 Joules, right?

That is, in your frame, you "transfer" 0.3 Joules of energy into the target.

Now. From the point of view of the target, what do they see? They see a laser pulse at 45 milliWatts, for 20 seconds, for a total of 0.9 Joules. The laser pulse is delivering photons three times more rapidly, but each photon is also blueshifted to hold three times as much energy. Power is up by a factor of 9. The duration, however, is reduced.

In their frame, you have "transferred" 0.9 Joules of energy from the laser.

Do you agree with these numbers?

There IS an invariant quantity, in all frames, of 0.9 Joules increase in the rest energy of the target. In the laser frame, the target reduces in velocity from the impact of the laser pulse, and because it is moving so fast, that is a substantial amount of energy that ends up absorbed into the frame of the target, adding to the damage done. THAT'S why it is NOT the "transferred" energy which is the constant.

Case one (laser frame). Laser loses 0.3 Joules of battery power, and picks up a negligible kinetic energy in recoil.

The target loses 0.6 Joules of kinetic energy, and gains 0.9 joules of heat.

Case two. (target frame) Laser loses 0.6 Joules of kinetic energy (recoil) and 0.3 Joules of battery, target gets 0.9 joules of heat, plus a negligible recoil from the impact of photons.

The transfer from one body to the other is different depending on the observer. But the heat taken in by the target is the same.

Cheers -- sylas

Last edited: May 28, 2009
18. May 28, 2009

### neopolitan

Okay, gotcha.

It's like there are three systems, photon/laser, photon/target and photon/laser/target and there is conservation in all three systems, right?

I have to admit I didn't take into account the recoil of the laser. Thanks for pointing it out sylas.

I see what Vanadium meant too though, there is no magic energy creation (or momentum creation).

cheers,

neopolitan

19. May 28, 2009

### sylas

Yes. In every inertial frame, there is conservation of energy, and conservation of momentum. In every frame, you get the same values for rest mass of the laser and of the target, before and after the laser burst. There's no free energy anywhere here.

What is different is the amount by which total energy changes for individual particles. That is, the quantity of energy and hence of momentum being transferred from the laser to the target is dependent on the observer. It scales precisely as the redshift or blueshift of the laser beam. It always balances, however, for any observer, with the net energy lost from the laser always equal to what is absorbed into the target.

Cheers -- sylas

20. May 28, 2009