3 ms·

> Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative

> Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above. I don't think this quite works because of relativistic addition of velocities. Naively, it seems to. For example, if the object was travelling at 0.999999c relative to you (appears to be 1.0c according to your limited instruments), then you accelerate to 0.50c in its direction, you'd see its speed reduce to 0.50c (same 2s.f. instrument), which would clearly prove it's not massless. But velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c because it only decreased a tiny amount, below what you instrument can detect. If you use a more precise instrument or a faster rocket, you might measure it as 1.00000c but then you still won't know if it's exactly c or a smidgen less. Maybe I've got my relativistic velocity addition wrong? But it still looks like the same measurement problem as trying to prove a classical object has a speed of exactly 0, which can't be done no matter how accurate our instruments are.

pdonis 6y ago
> velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c Not indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.
pdonis 6y ago
> Maybe I've got my relativistic velocity addition wrong? For the specific case you give, that depends on what accuracy you are assuming. The relativistic velocity addition would be: v_new = (0.999999 - 0.5) / (1 - 0.999999 * 0.5) = 0.999997. So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration. Also, suppose you accelerated for a second increment equal to the first; you would get v_new = (0.999997 - 0.5) / (1 - 0.999999 * 0.5) = 0.999991. And one more increment: v_new = (0.999991 - 0.5) / (1 - 0.999999 * 0.5) = 0.999973. As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.

pdonis 6y ago

> velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c Not indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.

pdonis 6y ago

> Maybe I've got my relativistic velocity addition wrong? For the specific case you give, that depends on what accuracy you are assuming. The relativistic velocity addition would be: v_new = (0.999999 - 0.5) / (1 - 0.999999 * 0.5) = 0.999997. So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration. Also, suppose you accelerated for a second increment equal to the first; you would get v_new = (0.999997 - 0.5) / (1 - 0.999999 * 0.5) = 0.999991. And one more increment: v_new = (0.999991 - 0.5) / (1 - 0.999999 * 0.5) = 0.999973. As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.