Parallel velocity hydrogen/antihydrogen annihilation at relativistic speeds.

danR

They've just now kept antihydrogen in a bottle for some 15 minutes.

Suppose we inject some H and measure the annihilation photon energies.

Then we accelerate the mixture (somehow) to a velocity around .99c and measure the energies of those photons that are emerging orthogonal to the velocity of our mixture. Are the energies the same as those resulting from the rest condition?

PAllen

Forget all the stuff about anti-hydrogen. In some frame you have light emitted in some direction. You ask, how does its frequency look in a frame moving at .99c orthogonal to the emission direction in the frame at rest with the emitter? That is just the relativistic transverse doppler. From the change in frequency, you get change in energy. I only took a quick glance at the following, but it looked reasonable:

http://en.wikipedia.org/wiki/Transve...Doppler_effect

danR

anti-H is not essential to the discussion, but any decay mode that produces photons.

Does an unstable particle that emits light generate (by measurement of the transversely emitted radiation) higher frequency light, shorter wavelength light, more energetic photons, in frame moving at .99c with respect to an external frame? Or are they same?

Does the averaged energy/wavelength of all the emitted photons in the same frame equal the averaged energy of all the photons when there is the .99c differential.

The motivation for the question involves decay energies, not merely any transverse differential of any emitter. I don't want to get into why, just yet.

DaleSpam

Two particles with four momenta (1,0,0,0) and (1,0,0,0) can annihilate to form a pair of photons with four momenta (1,0,1,0) and (1,0,-1,0). If you boost those four momenta in the x direction then the t components will be greater than 1, but they will not be orthogonal to the velocity. The orthogonality condition is not possible for both photons.

PAllen

In practice, this is rather complicated. The transverse doppler effect is of order v^2/c^2. The longitudinal doppler is of order v/c. Thus, if the you are measuring photons that are only approximately orthogonal to the relative motion between emitter and receiver, the normal doppler will swamp the transverse doppler. Thus, if you imagine a gamma ray emitter moving at .99c relative to a measuring device, the transverse doppler will be the dominant effect for only the tiniest moment.

You are being vague about experimental setup. To be precise, let's say you have an emitter of exactly 1 mev gamma rays, with very frequent emission, moving by a detector at .99c. As it approaches from the distance, the gamma rays received start out at about 14 mev, decreasing to 1 mev just before the emitter is exactly transverse (you will detect this photon when the emitter is just past transverse), then decreasing all the way to appx. 1/14 mev as the emitter recedes to the distance.

Hope this at least helps (at least for formulating a follow up question).

PAllen

As another example showing the importance of the experimental setup, imagine instead of the linear example I gave in my last post, you have:

A beam of highly excited nuclei in synchrotron at .99c. Let's again supposed they emit 1 mev gamma rays as they lose excitation, and you have a detector in the center of the ring. Now any gamma ray you receive would have pure transverse doppler red shift. You would get appx. 140 kev gammas at the central detector.

danR

OK, perhaps I can simplify this, and at the same time get to what I'm really driving at. We will dispense with the dry experimental setup. But excuse some unlikely science-fiction.

They have just trapped anti-H for some 15 minutes. 200 years from now, they have an H/anti-H drive spaceship.

The spaceship leaks a certain amount of radiation from its photonic drive through its infinitanium interleaved hyperdichroic mirror engine walls when it leaves earth. We won't worry anymore about direction. Consider all the photons.

When it flashes past Alpha Centauri at .999999c, is the average wavelength of all measurable photons <,>, or = that measured on Earth? I appreciate that the Centaurians have an understanding government in funding rather large photon detectors.

I appreciate that will take a lot of work to calculate, taking all relativistic effects into account.

Since a simplified version is what particle accelerator experimentalists deal with all the time, what is the standard and expected result for unstable particles involving photon emission as a decay product? I don't know if they've fired an H and -H together down an accelerator yet, but the theoretical answer would be satifactory also.

I will not be happy with emission from excited nuclei in an accelerator. Please humour me on that. I want particles that actually decay, or mutually annihilate, with photon emission. I definitely have a reason for that nuance.

PAllen

Please specify clearly who is measuring what photons. You have lot's of verbiage, but minimal specification. I get you have a rocket accelarating towards a star, passing it at very near c. Now specify who is measuring what photons. I don't see a clear specification of this in your post.

danR

Let's just say that you are measuring the photons. The photons emitted by annihilation of hydrogen and antihydrogen. If it will simplify your calculations, just use protons and antiprotons; I assume that will not matter. By 'measuring the photons' I mean measuring their energies. You are on a planet circling Alpha Centauri. It may be an imaginary planet, but that's immaterial. You are measuring all the photons. Don't worry about the transversely emitted photons. Just measure them all; all the photons emitted from the ship's engine. A colleague on Earth has measured the photons when the ship was in the relatively stationary reference frame on Earth, and will radio the results in due course for your comparison. You don't have to worry about the ship accelerating. Assume it has exhausted its fuel, and is moving past your vantage point (without any significant acceleration), that is to say, the vantage point of the detectors you have positioned around a point conveniently near your planet. Assume any orbital motions or relative motions of Earth and the Alpha Centauri planet are immaterial.

Now, that's the verbose version. If that's too verbose, I will condense it, but I suspect PF is not equipped to address the question, and I will have to take it to someone here at the university. Some of the antihydrogen team that recently trapped the antihydrogen for several minutes are researchers here.

danR

Never mind. This post belongs in the Particle Physics, I think. If not, I'll ask around here at the university, if there are any physics profs not on holidays.

PAllen

Verbosity isn't the issue. It is lack of specificiy. You start with "you are measuring the photons". That is completely ambiguous off the get go.

I will make a guess: Alpha cenauri is measuring anniliation photons from the rocket exhaust as it goes by. They are comparing the result to proton annililation in a local lab. In this case, my post #5 completely answers this question: as the rocket approaches, the photons will be received with (enormous) excess energy, decreasing to the same energy (for an instant) then rapidly decreasing in energy relative to lab photons as the rocket recedes.

So what is it your are really asking?

DaleSpam

PF already answered the question in post 4.

pervect

The original question is ambiguous, unfortunately. I'll interpret it in one particular way for definiteness.

If you set up a detector with a long tube, or a couple of holes in a plate, said assembly being at rest in the rest frame with a detector, you have one meaning for "orthogonal to the spaceship".

PIPE AND DETECTOR AT REST
detector ======(pipe)====== ^^^ spaceship (moving upwards)

In this case you'll see the "transverse doppler shift", or time dilation of 1/sqrt(1-v^2/c^2)

However, from the POV of the space=ship, the emitted light will not be "orthogonal to the spaceship". Here is the ambiguity in your question. In the frame of the spaceship, the emittted beam that passes through the pipe will be slanted significantly towards the rear.

danR

Post 4 is an excellent reply to a physics graduate. It's not one a layperson can put in a scrapbook or show to his friends, and say, for example: 'Ha, you see? I'm right and you're wrong.'

Or radio ahead to the Centaurians and say, 'Don't worry, we're just passing through, and the stray leakage photons won't vaporize your stellar system. We measured the leakage in static-tests here on Earth, and the levels are quite safe. It should be exactly the same when we fly past.'

I would want to be able to send that message with confidence.

DaleSpam

So then why didn't you ask questions about what you don't understand rather than make rude and unfounded comments about the capability of the posters on this site? I don't know you personally, so I have no way of knowing that you didn't understand if you don't ask for clarification.

Do you know what four-vectors are?