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If some kind of source was able to supply an infinite amount of energy, does that imply that it also must have an infinite mass? Is the contrary also true?

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Hello cipher. This is an imaginary question. Anyways, Have you taken $E=mc^2$ for that..? – Ϛѓăʑɏ βµԂԃϔ Oct 20 '12 at 13:49
Indeed i have. E=mc² is what actually made me ask this question – cipher Oct 20 '12 at 14:08
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@CrazyBuddy While you grammatical and tag edits were very good, please do not completely change a title like that without firm evidence that you are on the right track. – dmckee♦ Oct 20 '12 at 17:02
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The conjecture here is trivially true, but uninteresting because it is non-physical. – dmckee♦ Oct 20 '12 at 17:04

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It is very dangerous to talk about infinity in physics, especially when talking special relativity.

To your question: Yes the object would need an infinite amount of mass. E.G. Take a battery that would have an finite amount of energy inside.

Then you would have

$M_{\text{Total mass of the battery}}=M_{0,(\text{the usual Mass of an uncharged battery})}+\frac{E_{\text{Amount of Energy the Battery can supply}}}{c^2}$

Now If you set $\lim E\rightarrow \infty$ of course $M_{\text{Total mass of the battery}}$ also diverges

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For a photon $E = |p|c$. When $p \rightarrow \infty$ its energy $E \rightarrow \infty$ but its mass continues being zero.

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It's rest mass remains zero. But it's gravitational interactions will continue to be dictated by mass-energy just as always. – Colin K Oct 20 '12 at 22:29
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I dislike the term "rest mass"; specially for photons, which are never at rest. "Invariant mass" or simply "mass" are better terms. Gravitational interactions are given by the stress-energy-momentum tensor $T_{\mu\nu}$. Setting $m=0$ we obtain the form of the interaction for photons. – juanrga Oct 21 '12 at 11:01

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