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Einstein's energy equation (after substituting the equation of relativistic momentum) takes this form: $$E = \frac{1}{{\sqrt {1 - {v^2}/{c^2}} }}{m_0}{c^2} % $$ Now if you apply this form to a photon (I know this is controversial, in fact I would not do it, but I just want to understand the consequences), you get the following: $$E = \frac{1}{0}0{c^2}% $$ On another note, I understand that after dividing by zero:

  • If the numerator is any number other than zero, you get an "undefined" = no solution, because you are breaching mathematical rules.
  • If the numerator is zero, you get an "indeterminate" number = any value.

Here it seems we would have an "indeterminate" [if (1/0) times 0 equals 0/0], although I would prefer to have an "undefined" (because I think that applying this form to a photon breaches physical/logical rules, so I would like the outcome to breach mathematical rules as well...) and to support this I have read that if a subexpression is undefined (which would be the case here with gamma = 1/0), the whole expression becomes undefined (is this right and if so does it apply here?).

So what is the answer in strict mathematical terms: undefined or indeterminate?

Sierra
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    $0/0$ is both undefined and indeterminate. – ziggurism Dec 30 '17 at 20:41
  • You can define $0/0$ to be whatever you like. Whether that definition will be consistent within whatever axioms you are working in is another question altogether. – Project Book Dec 30 '17 at 20:56
  • Undefined and since Undefined $\implies$ indeterminate, we see that it is indeterminate too. – copper.hat Dec 30 '17 at 20:56
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    @copper.hat using the calculus terminology of indeterminate forms for limits, we would say that a form of type $1/0$ is not indeterminate (ratios of functions with those limits cannot take any possible values; they can only diverge). Yet $1/0$ is still undefined. So I am not sure that "undefined implies indeterminate" is correct, at least in this lingo. – ziggurism Dec 30 '17 at 21:02
  • If something is not defined then surely it is indeterminate. – copper.hat Dec 30 '17 at 21:05
  • On the physics side, the equation referenced applies only to particles with non-zero rest mass. The energy of a photon depends on the frequency of its associated electromagnetic field. – Mark Viola Dec 30 '17 at 21:22
  • @MarkViola That is why I said in my question that I think that applying the referenced equation to a photon is wrong, but I want to understand what happens, in technical mathematical terms, if you do such wrong thing. (If it is actually wrong at all, because respected physicist think otherwise and they are happy with the outcome: they say that it means that the photon does have "relativistic mass".) – Sierra Dec 30 '17 at 21:42

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I don't think you've quite come to terms yet with what undefined means. It means quite literally this expression has no meaning. So $1/0$ is literally meaningless even though it might look like it has meaning to our human brains, we haven't defined it, so it really has no meaning. For this reason, while we know that $x\cdot 0 = 0$ for any real number $x$, since $1/0$ is undefined, we don't even know that it is a real number, so we can't say $(1/0)\cdot 0 = 0$ nor can we say it is anything else for that matter. It is also undefined, because if we don't know what $1/0$ is, we can't know what anything made out of it is either.

Now of course one can try to extend definitions to define expressions that are currently undefined. However, for expressions such as the one you've given, the reason no one has done this is because there isn't a good way to do so.

Now the phrase "indeterminate form" is honestly not a great phrase. Either something has a value or it doesn't. Usually it's used to describe limits that don't obviously exist and/or whose value isn't obvious. For example while $$\lim_{x\to 0} \frac{\sin x}{x} = 1,$$ people often say that it has an indeterminate form, since $$\lim_{x\to 0} \sin x = 0 = \lim_{x\to 0} x.$$ Thus a naive approach to the limit would seem to give you that the limit is $0/0$, which is undefined. However, it turns out the limit is quite well-defined, it's just that the naive approach to taking limits of fractions doesn't work when the denominator goes to 0.

Point being, your expression is undefined, and I would not describe it as indeterminate (though of course this depends on your definitions and where you're from). As I was taught, the phrase indeterminate form (if used at all) should be reserved for well-defined expressions such as limits whose existence or values are not obvious because a naive approach to taking such limits results in an undefined expression.

jgon
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  • A very defined answer. – copper.hat Dec 30 '17 at 21:11
  • @jgon I had said that 1/0 = "undefined" means "no solution = you have breached the rules", but if you say it means "this is meaningless", I am also fine..., although I would prefer to stick to the idea that it entails a breach of logical rules, like if in common parlance you commit a fallacy, isn't that possible? – Sierra Dec 30 '17 at 21:46
  • I mean, if you'd like it's a breach of the rules, sure. What I was trying to get across is that $(1/0)\cdot 0$ is exactly as meaningful as $)\cdot 0$, or for those who need balanced parentheses, $()\cdot 0$. It is a breach of the rules, but I'm not sure it's a breach of logical rules or a fallacy, so much as a breach of the rules for forming valid expressions. – jgon Dec 30 '17 at 21:48
  • @jgon I had also said that I thought that "if a subexpression is undefined (which would be the case here with gamma = 1/0), the whole expression becomes undefined". You say "if we don't know what 1/0 is, we can't know what anything made out of it is either". Aren't the two statements equivalent? – Sierra Dec 30 '17 at 21:50
  • Yes, that statement is right. If any subexpression is undefined, the whole expression is undefined. – jgon Dec 30 '17 at 21:51
  • @jgon Last question and thanks a lot: what about 0/0? Depending on how we look at the above equation, this could perfectly be a reading thereof. But I gather from your clarifications (and looks quite logical to me) that the answer would not change: also if you are faced with a 0/0, you have an "undefined = a meaningless thing = something breaching the rules for forming valid expressions". Right? – Sierra Dec 30 '17 at 22:04
  • Yes 0/0 is undefined. – jgon Dec 30 '17 at 22:05
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Undefined is not a number. There is no such number as undefined, for which you could define the multiplication operation.

You could extend a set of a numbers (the set of the real numbers or the set of the complex numbers) with a new element, what you call "undefined", and then define a multiplication on this set as usual. It might be possible (although there are major problems solve).

However, from this moment, you have a defined value what you still call "undefined". Well, paper can hold everything, but it does not really sound as useful mathematics.

peterh
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  • Note, this is a mathematical answer. The physical answer is that there is no experimental evidence what would violate the Lorenz-invariance. All physical theories have their limit of validity. For example, you can say that water is a gas above 100 ${}^\circ$ C. You can't say what is water on $10^9 {}^\circ C$, because it is not water any more. Theories become typically invalid around their singularities (i.e. where formulas are giving division by zero). – peterh Sep 16 '19 at 16:50
  • Another note: although it does not work, thinking on things like this is imho useful. – peterh Sep 16 '19 at 16:54
  • In your example, the photons are described by different formulas. For example, they go always with $c$, and their impulse is proportional to their energy. No slower-than-light objects follow such laws. – peterh Sep 16 '19 at 16:58