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The other day somebody told me that there are absolute definitions of hot and cold for instance. You can always say a is warmer than b, no matter where you are. But he said, there is no absolute definition for right (i. e. not left). He told me that “they” came up with an experiment that you can set in front of you and something would go to the right. And if you turn the apparatus, it would still go to your right.

I said that I cannot believe this, since it would fail with two observers and right is just arbitrary. If he meant a definition of right in a chirality kind of way, anything with a cross product in it should allow him to separate right-hand from left-hand coordinate systems.

He again said that “they came up with an absolute definition of right. Just do not ask my how they did it”, which did not really allow for a substantial conversation.

Is there any such thing?

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I take it to be just the Righ-Hand-Rule via cross product... so if you move in direction $v_1$ and then turn 'to the right' in the orthogonal direction $v_2$ then $v_1\times v_2$ points 'upward', and this is independent of where you look. As for hot/cold, this is absolute because you can speak of "differences" between temperatures; i.e. you can shift the temperatures of each object by the same amount and their difference will be invariant... from this viewpoint, I don't know what you want "absolute" to mean for direction. –  Chris Gerig Dec 24 '11 at 23:24
    
Ultimately, I think your phrase "absolute definition" is ambiguous and you have to clarify it more to get something meaningful. –  Chris Gerig Dec 24 '11 at 23:24
    
the problem is usually stated as how do you communicate to an alien friend in another galaxy what right and left is. without any common reference this is impossible of course. it turns out that some particle reactions do not respect this symmetry and thus provide this needed common reference, and i'm sure this is what the friend is referencing. –  BjornW Dec 25 '11 at 0:47
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The practical way to define left and right is through weak decays of nuclei. If you take a nucleus which is unstable to $\beta$ decay, and you use a magnetic field to align the spin of the nucleus along the z-axis, the direction of charge flow in the coil generating the magnetic field defines a current loop. If you curl the fingers of your right hand along the direction of the current, the thumb points in the conventional direction of the magnetic field.

The electrons emitted during $\beta$ decay are emitted asymmetrically in the direction of the B axis. More electrons go one way than the other. The direction in which the electrons are preferentially emitted in each decay defines left and right in each $\beta$ decay experiment.

The reason for the asymmetry is ultimately because the neutrino only has one helicity, it only spins in a certain way around its direction of motion. So if you emit a neutrino/anti-neutrino in the down direction, you must lose/gain half a unit of spin in the x-y plane. Aligning the spin of the nucleus in a certain direction makes one direction easier than the other for neutrino emission.

This classic experiment was first performed by C.S. Wu in the late 1950s, after Lee and Yang suggested that parity is violated in the weak interaction. Sudarshan and Marshak, followed by Feynman and Gell-Mann, were the first to explain the phenomena by noting that the neutrino only has one helicity.

Right hand rule for EM is fake

The EM interaction respect parity (as do the strong interactions). But the description of EM elementary students get breaks parity, because it uses the right hand rule to define the magnetic field. The magnetic field does not break left-right symmetry, because you always use the right hand rule twice. Once to figure out the direction of the B field, and once to figure out the direction of the force from the B field.

It is possible to formulate EM without breaking the right hand rule, but at the cost of making B into an antisymmetric rank-2 tensor instead of an intuitive vector. So if a B field is pointing in the z-direction, the rank-2 tensor curls in the x-y plane.

But the chiral description of EM is actually understood to be more fundamental today, because of the magnetic monopoles required by quantum gravity. Magnetic monopoles break chiral symmetry for real. But we have no monopoles at low energy, so the undergraduate EM is not formulated with all the symmetry explicit.

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As for your last point, the RHR is perfectly fine, but it is the magnetic field that has the problem -- it is not a vector, but a pseudovector. –  Chris Gerig Dec 25 '11 at 5:16
    
@Chris Gerig: that's an alternate way of saying the same thing, but I prefer to say it with tensors. A pseudovector is the same as a two-index antisymmetric tensor, after mapping with the epsilon tensor, and the tensor description is much more natural. Tensor components naturally do not change sign under reflection, and the spatial part of the four dimensional Faraday tensor F is the antisymmetric magnetic field tensor. –  Ron Maimon Dec 25 '11 at 7:26
    
Okay, but the electrons would go to the other side if I rotate the experiment (with the magnetic field), right? So this definition of right and left is absolute in the ether sense either (which my friend suggests, which I do not believe though). –  queueoverflow Dec 25 '11 at 13:00
    
@queueoverflow: Hmm? Did you want a definition of left/right or of absolute direction? This experiment distinguishes between your right and left hand. There is no way to distinguish "up" from "down" or "left" from "right" in an absolute sense of picking out a direction. You can only distinguish between a shape and its mirror image. –  Ron Maimon Dec 25 '11 at 13:33
    
@RonMaimon: Okay, that is what I would think as well. –  queueoverflow Jan 27 '12 at 13:48
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I think what your frien talked about was chiral symmetry breaking, which was observe in particle physics experiments. In this sense, "right" and "left" may be absolutely defined, much like how positive and negative charges are absolutely defined although the names are arbitrary.

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-1: No, not Chiral symmetry breaking! Chiral symmetry breaking is left-right invariant. The parity violating weak interactions are the only way to distinguish left from right. QCD with chiral symmetry breaking is parity invariant. The name is slightly misleading--- the reason it is called that is because the chiral condensate breaks the chiral part of isospin, the part that does isospin rotations on the different chiralities of quarks in opposite directions. But parity just reflects the chiralities, making the chiral isospin gauge boson a pseudovector A. Parity is still well defined. –  Ron Maimon Dec 25 '11 at 4:44
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